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Amplify-and-<b>forward</b>

道理特分解法

#include "iostream" using namespace std; class Matrix { private: double** A; //矩陣A double *b; //向量b public: int size; Matrix(int ); ~Matrix(); friend double* Dooli(Matrix& ); void Input(); void Disp(); }; Matrix::Matrix(int x) { size=x; //為向量b分配空間并初始化為0 b=new double [x]; for(int j=0;j<x;j++) b[j]=0; //為向量A分配空間并初始化為0 A=new double* [x]; for(int i=0;i<x;i++) A[i]=new double [x]; for(int m=0;m<x;m++) for(int n=0;n<x;n++) A[m][n]=0; } Matrix::~Matrix() { cout<<"正在析構中~~~~"<<endl; delete b; for(int i=0;i<size;i++) delete A[i]; delete A; } void Matrix::Disp() { for(int i=0;i<size;i++) { for(int j=0;j<size;j++) cout<<A[i][j]<<" "; cout<<endl; } } void Matrix::Input() { cout<<"請輸入A:"<<endl; for(int i=0;i<size;i++) for(int j=0;j<size;j++){ cout<<"第"<<i+1<<"行"<<"第"<<j+1<<"列:"<<endl; cin>>A[i][j]; } cout<<"請輸入b:"<<endl; for(int j=0;j<size;j++){ cout<<"第"<<j+1<<"個:"<<endl; cin>>b[j]; } } double* Dooli(Matrix& A) { double *Xn=new double [A.size]; Matrix L(A.size),U(A.size); //分別求得U，L的第一行與第一列 for(int i=0;i<A.size;i++) U.A[0][i]=A.A[0][i]; for(int j=1;j<A.size;j++) L.A[j][0]=A.A[j][0]/U.A[0][0]; //分別求得U，L的第r行，第r列 double temp1=0,temp2=0; for(int r=1;r<A.size;r++){ //U for(int i=r;i<A.size;i++){ for(int k=0;k<r-1;k++) temp1=temp1+L.A[r][k]*U.A[k][i]; U.A[r][i]=A.A[r][i]-temp1; } //L for(int i=r+1;i<A.size;i++){ for(int k=0;k<r-1;k++) temp2=temp2+L.A[i][k]*U.A[k][r]; L.A[i][r]=(A.A[i][r]-temp2)/U.A[r][r]; } } cout<<"計算U得:"<<endl; U.Disp(); cout<<"計算L的:"<<endl; L.Disp(); double *Y=new double [A.size]; Y[0]=A.b[0]; for(int i=1;i<A.size;i++ ){ double temp3=0; for(int k=0;k<i-1;k++) temp3=temp3+L.A[i][k]*Y[k]; Y[i]=A.b[i]-temp3; } Xn[A.size-1]=Y[A.size-1]/U.A[A.size-1][A.size-1]; for(int i=A.size-1;i>=0;i--){ double temp4=0; for(int k=i+1;k<A.size;k++) temp4=temp4+U.A[i][k]*Xn[k]; Xn[i]=(Y[i]-temp4)/U.A[i][i]; } return Xn; } int main() { Matrix B(4); B.Input(); double *X; X=Dooli(B); cout<<"~~~~解得:"<<endl; for(int i=0;i<B.size;i++) cout<<"X["<<i<<"]:"<<X[i]<<" "; cout<<endl<<"呵呵呵呵呵"; return 0; }

標簽： 道理特分解法

上傳時間： 2018-05-20

上傳用戶：Aa123456789
基于數(shù)字電路的八路搶答器的設計與實現(xiàn)

搶答器是一種智力競賽常用的器件,搶答器的設計方法千差萬別,文章利用常用的數(shù)字電子器件,設計了八路搶答器電路的設計、仿真及實現(xiàn)的全過程,提出兩種可行的設計方案:方案1采用74ls373實現(xiàn)電路鎖存,74ls148實現(xiàn)電路編碼,74ls74及數(shù)碼管實現(xiàn)電路顯示;方案二采用CD4511BCN和LMC555CM集成電路及數(shù)碼管實現(xiàn)搶答器的控制和顯示。本文設計用的器件簡單,容易理解,適用于初學電子技術的人員。Answer scrambler is a common device in intelligence competition, and its design methods vary greatly. This paper designs the whole process of design, simulation and Realization of the circuit of eight-way answer scrambler by using common digital electronic devices, and puts forward two feasible design schemes: scheme 1 uses 74 ls373 to realize circuit latching, 74 ls148 to realize circuit coding,74 ls74 and digital tube to realize circuit. The second scheme uses CD4511 BCN, LMC555 CM integrated circuit and digital tube to control and display the answerer. The device designed in this paper is simple and easy to understand, and it is suitable for the beginners of electronic technology.

標簽： 搶答器

上傳時間： 2022-04-05

上傳用戶：
安森美車規(guī)級1080P圖像傳感器AR0231手冊

AR0231AT7C00XUEA0-DRBR（RGB濾光）安森美半導體推出采用突破性減少LED閃爍 (LFM)技術的新的230萬像素CMOS圖像傳感器樣品AR0231AT，為汽車先進駕駛輔助系統(tǒng)(ADAS)應用確立了一個新基準。新器件能捕獲1080p高動態(tài)范圍(HDR)視頻，還具備支持汽車安全完整性等級B(ASIL B)的特性。LFM技術(專利申請中)消除交通信號燈和汽車LED照明的高頻LED閃爍，令交通信號閱讀算法能于所有光照條件下工作。AR0231AT具有1/2.7英寸(6.82 mm)光學格式和1928(水平) x 1208(垂直)有源像素陣列。它采用最新的3.0微米背照式(BSI)像素及安森美半導體的DR-Pix?技術，提供雙轉(zhuǎn)換增益以在所有光照條件下提升性能。它以線性、HDR或LFM模式捕獲圖像，并提供模式間的幀到幀情境切換。 AR0231AT提供達4重曝光的HDR，以出色的噪聲性能捕獲超過120dB的動態(tài)范圍。AR0231AT能同步支持多個攝相機，以易于在汽車應用中實現(xiàn)多個傳感器節(jié)點，和通過一個簡單的雙線串行接口實現(xiàn)用戶可編程性。它還有多個數(shù)據(jù)接口，包括MIPI(移動產(chǎn)業(yè)處理器接口)、并行和HiSPi(高速串行像素接口)。其它關鍵特性還包括可選自動化或用戶控制的黑電平控制，支持擴頻時鐘輸入和提供多色濾波陣列選擇。封裝和現(xiàn)狀：AR0231AT采用11 mm x 10 mm iBGA-121封裝，現(xiàn)提供工程樣品。工作溫度范圍為-40℃至105℃(環(huán)境溫度)，將完全通過AEC-Q100認證。

標簽： 圖像傳感器

上傳時間： 2022-06-27

上傳用戶：XuVshu
A Simple isochronous transfer. Reads 8051 ports A,B and C, and continuously sends a five byte packet

A Simple isochronous transfer. Reads 8051 ports A,B and C, and continuously sends a five byte packet over EP8IN:

標簽： continuously isochronous and transfer

上傳時間： 2015-02-01

上傳用戶：ywqaxiwang
If we have two individually sorted vectors "a" and "b" but they are not sorted with respect to each

If we have two individually sorted vectors "a" and "b" but they are not sorted with respect to each other and we want to merge them into vector "c" such that "c" is also a sorted vector. Then c=mergesorted(a,b) can be used.

標簽： sorted individually respect vectors

上傳時間： 2015-09-23

上傳用戶：comua
Distributed Median,Alice has an array A, and Bob has an array B. All elements in A and B are distinc

Distributed Median,Alice has an array A, and Bob has an array B. All elements in A and B are distinct. Alice and Bob are interested in finding the median element of their combined arrays.

標簽： array B. Distributed has

上傳時間： 2013-12-25

上傳用戶：洛木卓
design LP,HP,B S digital Butterworth and Chebyshev filter. All array has been specified internally

design LP,HP,B S digital Butterworth and Chebyshev filter. All array has been specified internally,so user only need to input f1,f2,f3,f4,fs(in hz), alpha1,alpha2(in db) and iband (to specify the type of to design). This program output hk(z)=bk(z)/ak(z),k=1,2,..., ksection and the freq.

標簽： Butterworth internally Chebyshev specified

上傳時間： 2015-11-08

上傳用戶：253189838
mpeg test inter b. Dequantizer Algorithm hardware realization method and comparison c. Dequantize

mpeg test inter b. Dequantizer Algorithm hardware realization method and comparison c. Dequantizer Hardware Architecture Design

標簽： b. c. Dequantizer realization

上傳時間： 2016-01-29

上傳用戶：450976175
bool 運算(a)and(b)二進制十進制顯示

bool 運算(a)and(b)二進制十進制顯示

標簽： bool and 運算二進制

上傳時間： 2013-11-28

上傳用戶：waitingfy
TMS320C6000 DSP Power-Down Logic and Modes Reference Guide (Rev. B).pdf

TMS320C6000 DSP Power-Down Logic and Modes Reference Guide (Rev. B).pdf

標簽： Power-Down Reference C6000 Logic

上傳時間： 2014-12-07

上傳用戶：woshini123456